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Proof step 40 from proof diag-theorem-butlast (previous step)

# nextl(0, a) = ii(1) + 1, lf(n0 + 1) = nextl(n0, lf(n0).last), x to ii(1) + 1 = nextl(0, a), ii(0) = 0, lf(0) = a ', lf(1) = nextl(0, a), nextl(0, a) = a ' + y, ii(n0 + 1) = ii(n0) + # lf(n0 + 1) - 1, (x from ii(n0)) to # lf(n0 + 1) = lf(n0 + 1), 1 + ii(n) = # x, lf(m + 1) = nextl(m, lf(m).last), a ∈ nextl(0, a), ii(1) ≤ ii(n), ii(n0 + 1) ≤ ii(n), n ≤ m, ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ n. lf(n + 1) = nextl(n, lf(n).last) ∧ lf(n) ≠ [], ∀ n, a. nextl(n, a) ≠ [] ∧ nextl(n, a).first = a, ∀ m. n ≤ m → lf(m + 1).rest = [] ∧ ii(m) = ii(n), x[ii(m)] = lf(m).last, ∀ n1. ii(n1 + 1) = ii(n1) + # lf(n1 + 1) - 1 ∧ ¬ ii(n) < ii(n1 + 1) ∧ (x from ii(n1)) to # lf(n1 + 1) = lf(n1 + 1) ⊦ nextl(0, a) = [], lf(n0) = [], x = [], lf(m) = [], nextl(m, x[ii(m)]) = x[ii(m)] ' ∧ ii(n) = ii(m)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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